Question

Use the elimination method to determine whether the linear-quadratic system y=x2−4x+8
, y=3x−4
has zero, one, or two solutions.

Enter 0 for zero solutions.

Enter 1 for one solution.

Enter 2 for two solutions.

(1 point)
This system has
solution(s).

Answer 1

To determine the number of solutions for the system of equations

1. \( y = x^2 - 4x + 8 \) (quadratic equation)
2. \( y = 3x - 4 \) (linear equation)

we can set the two equations equal to each other since they both equal \( y \):

\[
x^2 - 4x + 8 = 3x - 4
\]

Now, let's rearrange the equation:

\[
x^2 - 4x - 3x + 8 + 4 = 0
\]

This simplifies to:

\[
x^2 - 7x + 12 = 0
\]

Next, we can factor the quadratic equation:

\[
(x - 3)(x - 4) = 0
\]

Setting each factor to zero gives us the potential solutions for \( x \):

1. \( x - 3 = 0 \) → \( x = 3 \)
2. \( x - 4 = 0 \) → \( x = 4 \)

Since we found two distinct values for \( x \), we will have two corresponding \( y \) values by substituting \( x \) back into either of the original equations.

Thus, the system has **2** solutions.

The answer is:

**2** solution(s).